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10.13 - Radius and Interval of Convergence of Power Series

What is a Power Series?

A power series is an infinite series of the form:
∑n=0∞anxn=a0+a1x+a2x2+a3x3+⋯\sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots
or more generally, centered at x=cx = c:
∑n=0∞an(x−c)n=a0+a1(x−c)+a2(x−c)2+a3(x−c)3+⋯\sum_{n=0}^{\infty} a_n (x-c)^n = a_0 + a_1(x-c) + a_2(x-c)^2 + a_3(x-c)^3 + \cdots
The center cc is the value the series is built around.
Identifying the center: - ∑anxn\sum a_n x^n → centered at c=0c = 0 - ∑an(x−5)n\sum a_n (x-5)^n → centered at c=5c = 5 - ∑an(x+3)n\sum a_n (x+3)^n → centered at c=−3c = -3
Important: the center always converges. At a bare minimum, the series converges at x=cx = c.

Three Types of Convergence for Power Series

A power series can converge in one of three ways.

Type 1: Converges on an Interval

The series converges for all xx in some interval around cc.
10.13 - Radius and Interval of Convergence of Power Series
The radius of convergence RR is the distance from the center cc to either endpoint:
R=∣b−c∣=∣c−a∣R = |b - c| = |c - a|

Type 2: Converges for All Real Numbers

The series converges for every value of xx.
Interval: (−∞,∞)(-\infty, \infty). Radius: R=∞R = \infty.

Type 3: Converges Only at the Center

The series converges only when x=cx = c, and diverges everywhere else.
Interval: just the single point {c}\{c\}. Radius: R=0R = 0.

Using the Ratio Test to Find Convergence

We apply the ratio test to find the interval of convergence:
L=lim⁡n→∞∣an+1an∣L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|

Three Cases Based on the Limit

Case 1: L<1L < 1
The series converges on an interval around cc. You have to check the endpoints separately to see if they’re included.
Case 2: L=0L = 0
The series converges for all xx, so the radius is R=∞R = \infty.
Case 3: L=∞L = \infty
The series converges only at the center x=cx = c, so the radius is R=0R = 0.

Step-by-Step Process

  1. Identify the center cc from the series
  1. Apply the ratio test: calculate lim⁡n→∞∣an+1an∣\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|
  1. Determine the case:
      • If the limit is <1< 1, solve the inequality for xx
      • If the limit is 00, it converges for all xx
      • If the limit is ∞\infty, it converges only at cc
  1. For Case 1 only, check the endpoints. Plug each endpoint into the original series and test for convergence
  1. Write the interval of convergence with the correct brackets and parentheses
  1. Find the radius: the distance from the center to an endpoint

Example 1: Interval with Endpoint Checking

Find the interval and radius of convergence for ∑n=1∞n(x−5)n3n\sum_{n=1}^{\infty} \frac{n(x-5)^n}{3^n}.
Step 1: Identify the center
c=5c = 5
Step 2: Apply the ratio test
an=n(x−5)n3n,an+1=(n+1)(x−5)n+13n+1a_n = \frac{n(x-5)^n}{3^n}, \qquad a_{n+1} = \frac{(n+1)(x-5)^{n+1}}{3^{n+1}}L=lim⁡n→∞∣an+1an∣=lim⁡n→∞∣(n+1)(x−5)n+13n+1⋅3nn(x−5)n∣L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left|\frac{(n+1)(x-5)^{n+1}}{3^{n+1}} \cdot \frac{3^n}{n(x-5)^n}\right|=lim⁡n→∞∣n+1n⋅x−53∣=1⋅∣x−53∣=∣x−53∣= \lim_{n \to \infty} \left|\frac{n+1}{n} \cdot \frac{x-5}{3}\right| = 1 \cdot \left|\frac{x-5}{3}\right| = \left|\frac{x-5}{3}\right|
Step 3: Solve L<1L < 1
∣x−53∣<1\left|\frac{x-5}{3}\right| < 1−1<x−53<1-1 < \frac{x-5}{3} < 1
Multiply by 3:
−3<x−5<3-3 < x-5 < 3
Add 5:
2<x<82 < x < 8
Step 4: Check the endpoints
At x=2x = 2:
∑n=1∞n(2−5)n3n=∑n=1∞n(−3)n3n=∑n=1∞n⋅(−1)n⋅3n3n=∑n=1∞n(−1)n\sum_{n=1}^{\infty} \frac{n(2-5)^n}{3^n} = \sum_{n=1}^{\infty} \frac{n(-3)^n}{3^n} = \sum_{n=1}^{\infty} \frac{n \cdot (-1)^n \cdot 3^n}{3^n} = \sum_{n=1}^{\infty} n(-1)^n
The terms are −1,2,−3,4,−5,6,…-1, 2, -3, 4, -5, 6, \ldots, which don’t approach zero. So this diverges by the nth term test.
At x=8x = 8:
∑n=1∞n(8−5)n3n=∑n=1∞n⋅3n3n=∑n=1∞n\sum_{n=1}^{\infty} \frac{n(8-5)^n}{3^n} = \sum_{n=1}^{\infty} \frac{n \cdot 3^n}{3^n} = \sum_{n=1}^{\infty} n
That’s 1+2+3+4+⋯1 + 2 + 3 + 4 + \cdots, which diverges.
Step 5: Interval of convergence
(2,8)(2, 8)
An open interval, since both endpoints are excluded.
Step 6: Radius of convergence
R=8−5=3orR=5−2=3R = 8 - 5 = 3 \quad \text{or} \quad R = 5 - 2 = 3

Example 2: Series with Different Endpoint Behavior

Find the interval and radius of convergence for ∑n=1∞3(x−2)nn⋅3n\sum_{n=1}^{\infty} \frac{3(x-2)^n}{n \cdot 3^n}.
Step 1: The center is c=2c = 2.
Step 2: Apply the ratio test
L=lim⁡n→∞∣3(x−2)n+1(n+1)⋅3n+1⋅n⋅3n3(x−2)n∣=lim⁡n→∞∣nn+1⋅x−23∣=∣x−23∣L = \lim_{n \to \infty} \left|\frac{3(x-2)^{n+1}}{(n+1) \cdot 3^{n+1}} \cdot \frac{n \cdot 3^n}{3(x-2)^n}\right| = \lim_{n \to \infty} \left|\frac{n}{n+1} \cdot \frac{x-2}{3}\right| = \left|\frac{x-2}{3}\right|
Step 3: Solve L<1L < 1
∣x−23∣<1\left|\frac{x-2}{3}\right| < 1−1<x−23<1-1 < \frac{x-2}{3} < 1−3<x−2<3-3 < x-2 < 3−1<x<5-1 < x < 5
Step 4: Check the endpoints
At x=−1x = -1:
∑n=1∞3(−1−2)nn⋅3n=∑n=1∞3(−3)nn⋅3n=∑n=1∞3(−1)nn\sum_{n=1}^{\infty} \frac{3(-1-2)^n}{n \cdot 3^n} = \sum_{n=1}^{\infty} \frac{3(-3)^n}{n \cdot 3^n} = \sum_{n=1}^{\infty} \frac{3(-1)^n}{n}
That simplifies to 3∑n=1∞(−1)nn3 \sum_{n=1}^{\infty} \frac{(-1)^n}{n}, which is 3 times the alternating harmonic series. The alternating harmonic series converges by the alternating series test, so this converges.
At x=5x = 5:
∑n=1∞3(5−2)nn⋅3n=∑n=1∞3⋅3nn⋅3n=∑n=1∞3n\sum_{n=1}^{\infty} \frac{3(5-2)^n}{n \cdot 3^n} = \sum_{n=1}^{\infty} \frac{3 \cdot 3^n}{n \cdot 3^n} = \sum_{n=1}^{\infty} \frac{3}{n}
That’s 3 times the harmonic series, which diverges.
Step 5: Interval of convergence
[−1,5)[-1, 5)
The left endpoint is included, the right one isn’t.
Step 6: Radius
R=5−2=3orR=2−(−1)=3R = 5 - 2 = 3 \quad \text{or} \quad R = 2 - (-1) = 3

Example 3: Both Endpoints Included

Find the interval and radius of convergence for ∑n=1∞(x+2)nn3\sum_{n=1}^{\infty} \frac{(x+2)^n}{n^3}.
Step 1: The center is c=−2c = -2.
Step 2: Apply the ratio test
L=lim⁡n→∞∣(x+2)n+1(n+1)3⋅n3(x+2)n∣=lim⁡n→∞(nn+1)3∣x+2∣=1⋅∣x+2∣=∣x+2∣L = \lim_{n \to \infty} \left|\frac{(x+2)^{n+1}}{(n+1)^3} \cdot \frac{n^3}{(x+2)^n}\right| = \lim_{n \to \infty} \left(\frac{n}{n+1}\right)^3 |x+2| = 1 \cdot |x+2| = |x+2|
Step 3: Solve L<1L < 1
∣x+2∣<1|x+2| < 1−1<x+2<1-1 < x+2 < 1−3<x<−1-3 < x < -1
Step 4: Check the endpoints
At x=−3x = -3:
∑n=1∞(−3+2)nn3=∑n=1∞(−1)nn3\sum_{n=1}^{\infty} \frac{(-3+2)^n}{n^3} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}
That’s an alternating series with terms 1n3\frac{1}{n^3} decreasing to zero, so it converges by the alternating series test.
At x=−1x = -1:
∑n=1∞(−1+2)nn3=∑n=1∞1n3\sum_{n=1}^{\infty} \frac{(-1+2)^n}{n^3} = \sum_{n=1}^{\infty} \frac{1}{n^3}
That’s a p-series with p=3>1p = 3 > 1, so it converges.
Step 5: Interval of convergence
[−3,−1][-3, -1]
A closed interval, since both endpoints are included.
Step 6: Radius
R=−1−(−2)=1orR=−2−(−3)=1R = -1 - (-2) = 1 \quad \text{or} \quad R = -2 - (-3) = 1

Example 4: Converges Only at the Center (L=∞L = \infty)

Find the interval and radius of convergence for ∑n=0∞(2n)!⋅(x−1)2nn!\sum_{n=0}^{\infty} \frac{(2n)! \cdot (x-1)^{2n}}{n!}.
Step 1: The center is c=1c = 1.
Step 2: Apply the ratio test
L=lim⁡n→∞∣(2n+2)!⋅(x−1)2n+2(n+1)!⋅n!(2n)!⋅(x−1)2n∣L = \lim_{n \to \infty} \left|\frac{(2n+2)! \cdot (x-1)^{2n+2}}{(n+1)!} \cdot \frac{n!}{(2n)! \cdot (x-1)^{2n}}\right|
Simplify, using (2n+2)!=(2n+2)(2n+1)(2n)!(2n+2)! = (2n+2)(2n+1)(2n)! and (n+1)!=(n+1)⋅n!(n+1)! = (n+1) \cdot n!:
=lim⁡n→∞∣(2n+2)(2n+1)(2n)!(n+1)n!⋅n!(2n)!⋅(x−1)2∣= \lim_{n \to \infty} \left|\frac{(2n+2)(2n+1)(2n)!}{(n+1)n!} \cdot \frac{n!}{(2n)!} \cdot (x-1)^2\right|=lim⁡n→∞(2n+2)(2n+1)n+1⋅∣x−1∣2= \lim_{n \to \infty} \frac{(2n+2)(2n+1)}{n+1} \cdot |x-1|^2=lim⁡n→∞4n2+6n+2n+1⋅∣x−1∣2=∞⋅∣x−1∣2=∞= \lim_{n \to \infty} \frac{4n^2 + 6n + 2}{n+1} \cdot |x-1|^2 = \infty \cdot |x-1|^2 = \infty
For any x≠1x \neq 1, this goes to infinity.
Step 3: Since L=∞L = \infty, the series converges only at the center.
Interval: {1}\{1\}, just the single point x=1x = 1. Radius: R=0R = 0.

Example 5: Converges for All Real Numbers (L=0L = 0)

Find the interval and radius of convergence for ∑n=0∞xnn!\sum_{n=0}^{\infty} \frac{x^n}{n!}.
Step 1: The center is c=0c = 0.
Step 2: Apply the ratio test
L=lim⁡n→∞∣xn+1(n+1)!⋅n!xn∣=lim⁡n→∞∣x∣n+1=0L = \lim_{n \to \infty} \left|\frac{x^{n+1}}{(n+1)!} \cdot \frac{n!}{x^n}\right| = \lim_{n \to \infty} \frac{|x|}{n+1} = 0
Step 3: Since L=0L = 0, the series converges for all xx.
Interval: (−∞,∞)(-\infty, \infty). Radius: R=∞R = \infty.
Note: this is the Maclaurin series for exe^x.

Important Notes on Checking Endpoints

When to Check Endpoints

Only when L<1L < 1 gives you a finite interval.
Don’t check endpoints when: - L=0L = 0, since it converges everywhere - L=∞L = \infty, since it converges only at the center

How to Check Endpoints

Substitute the endpoint value into the original series and test it using: - The nth term test, since if the terms don’t approach zero it diverges - The p-series test, if it becomes ∑1np\sum \frac{1}{n^p} - The alternating series test, if it alternates with decreasing terms - The geometric series test, if it becomes ∑arn\sum ar^n - Any other convergence test that fits

Writing the Interval Correctly

  • Both converge: [a,b][a, b]
  • Left converges, right diverges: [a,b)[a, b)
  • Left diverges, right converges: (a,b](a, b]
  • Both diverge: (a,b)(a, b)

Finding the Radius Without Finding the Interval

If the problem only asks for the radius, you can skip checking the endpoints. Just solve L<1L < 1 and find the distance from the center to the edge.
Shortcut: once you’ve solved the inequality, the radius is:
R=∣either endpoint−c∣R = |\text{either endpoint} - c|

What a Power Series Represents

Inside its interval of convergence, a power series adds up to one number for each xx, so it defines a function: f(x)=∑an(x−c)nf(x) = \sum a_n (x-c)^n.
Key fact: if a power series has a positive radius of convergence (including R=∞R = \infty), then on the open interval (c−R,c+R)(c - R, c + R) it is the Taylor series for the function it converges to. Its coefficients have to be
an=f(n)(c)n!a_n = \frac{f^{(n)}(c)}{n!}
So a function has only one power series centered at cc. However you build it (derivatives, substitution, a geometric series), you get the same Taylor series, and you can read derivatives right off it: f(n)(c)=n!⋅anf^{(n)}(c) = n! \cdot a_n.
Example: the geometric series ∑n=0∞xn\sum_{n=0}^{\infty} x^n converges to f(x)=11−xf(x) = \frac{1}{1-x} on (−1,1)(-1, 1). Its radius is R=1>0R = 1 > 0, so it’s the Maclaurin series for 11−x\frac{1}{1-x}. Every coefficient is 11, so f(n)(0)=n!⋅1=n!f^{(n)}(0) = n! \cdot 1 = n!. (Check: f′(x)=(1−x)−2f'(x) = (1-x)^{-2} and f′′(x)=2(1−x)−3f''(x) = 2(1-x)^{-3}, so f′(0)=1f'(0) = 1 and f′′(0)=2=2!f''(0) = 2 = 2!.)

Differentiating or Integrating Keeps the Radius

If you differentiate or integrate a power series term by term (10.15), the new series has the same radius of convergence. The endpoints can change, though, so check them again.
  • ∑n=1∞xnn\sum_{n=1}^{\infty} \frac{x^n}{n} converges on [−1,1)[-1, 1): at x=−1x = -1 it’s the alternating harmonic series, and at x=1x = 1 it’s the harmonic series
  • Its derivative, ∑n=1∞xn−1=1+x+x2+⋯\sum_{n=1}^{\infty} x^{n-1} = 1 + x + x^2 + \cdots, converges on (−1,1)(-1, 1): at both endpoints the terms don’t go to 0
  • Its integral from 0, ∑n=1∞xn+1n(n+1)\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}, converges on [−1,1][-1, 1]: at both endpoints the terms are no bigger than 1n2\frac{1}{n^2} in absolute value
Same radius every time (R=1R = 1), three different intervals.

Summary Table of Cases

Limit Value
Convergence
Interval
Radius
L<1L < 1
On an interval
Solve the inequality, check endpoints
Distance from cc to an endpoint
L=0L = 0
All real numbers
(−∞,∞)(-\infty, \infty)
R=∞R = \infty
L=∞L = \infty
Only at the center
{c}\{c\}
R=0R = 0

Common Mistakes to Avoid

  • ❌ Confusing L=0L = 0 with L=∞L = \infty. L=0L = 0 means it converges for all xx (infinite radius), while L=∞L = \infty means it converges only at the center (zero radius)
  • ❌ Forgetting to check the endpoints. You have to test each one separately when L<1L < 1
  • ❌ Forgetting the absolute value in the ratio test. Always use ∣an+1an∣\left|\frac{a_{n+1}}{a_n}\right|
  • ❌ Using the wrong inequality. Set L<1L < 1, not L≤1L \leq 1
  • ❌ Not identifying the center correctly. (x−5)(x-5) means the center is +5+5, and (x+3)(x+3) means the center is −3-3
  • ❌ Miscalculating endpoint convergence. Substitute and simplify carefully before testing

Key Takeaways

  • Power series: ∑an(x−c)n\sum a_n(x-c)^n, centered at x=cx = c
  • The center always converges. At a minimum, the series converges at x=cx = c
  • Use the ratio test: L=lim⁡n→∞∣an+1an∣L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|
  • Three cases: L<1L < 1 (an interval), L=0L = 0 (all xx), L=∞L = \infty (only the center)
  • Check the endpoints separately when L<1L < 1, to determine whether the interval is open or closed
  • Radius = the distance from the center to the edge of the convergence interval
  • If L=0L = 0: it converges for all xx, and the radius is R=∞R = \infty
  • If L=∞L = \infty: it converges only at cc, and the radius is R=0R = 0
  • Endpoint tests: use the nth term test, p-series, alternating series, or whatever fits
  • Interval notation: [  ][\;] for included endpoints, (  )(\;) for excluded ones
  • Radius only? You don’t need to check the endpoints if only the radius is asked for
  • Positive radius: the series is the Taylor series of the function it converges to, and term-by-term derivatives or integrals keep the same radius (recheck the endpoints)
  • The series ∑xnn!=ex\sum \frac{x^n}{n!} = e^x converges for all xx, with radius ∞\infty

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